If 2x – 3y = 7 and (a + b)x – (a + b – 3)y = 4a + b have an infinite number of solutions then

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a = –5, b = –1 Explanation: 1. Identify system coefficients For a pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, an infinite number of solutions exists if and only if their corresponding ratios are equal: `(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)` Rewriting the given equations into standard form: Equation 1: 2x – 3y – 7 = 0 $\Rightarrow$ a1 = 2, b1 = –3, c1 = –7 Equation 2: (a + b)x – (a + b – 3)y – (4a + b) = 0 $\Rightarrow$ a2 = a + b, b2 = –(a + b – 3), c2 = –(4a + b) Substituting these values into the ratio condition givesTgQPHd||[]: `2/(a + b) = (-3)/(-(a + b - 3)) = (-7)/(-(4a + b)) $\Rightarrow$ 2/(a + b) = (-3)/(a + b - 3) = (7)/(4a + b)` 2. Set up equations Equating the first two ratios to find our first relation: 2/(a + b) = 3/(a + b - 3) Cross-multiplying: 2(a + b – 3) = 3(a + b) 2a + 2b – 6 = 3a + 3b a + b = –6 ...(Equation I) Equating the first and third ratios to find our second relation: 2/(a + b) = 7/(4a + b) Cross-multiplying: 2(4a + b) = 7(a + b) 8a + 2b = 7a + 7b a = 5b ...(Equation II) 3. Solve for variables Substitute Equation II into Equation I: 5b + b = –6 6b = –6 $\Rightarrow$ b = –1 Now substitute b = –1 back into Equation II: a = 5(–1) $\Rightarrow$ a = –5

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